Recall a commutative ring $R$ is called a reduced ring if it has no non-zero nilpotent elements and a minimal ideal of $R$ is a non-zero ideal which contains no other nonzero right ideal. Clearly the annihilator of minimal ideal is maximal and two minimal ideals are isomorphic as $R$-moduls if and only if they have the same annihilators.

How can we show that in a commutative reduced ring minimal ideals are non-isomorphic?

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    $\begingroup$ I have improved the quotation. $\endgroup$
    – A.Rajanda
    Feb 26, 2018 at 5:16
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    $\begingroup$ If $I$ is a minimal ideal of a ring $R$, then $I\cong R/ann(I)$ as $R$-modules. Hence, $R/ann(I)$ is a simple module and so $ann(I)$ must be maximal. $\endgroup$
    – A.Rajanda
    Feb 27, 2018 at 6:08
  • $\begingroup$ sorry, right. I was thinking of a different problem: the annihilator of a maximal isn’t necessarily minimal. $\endgroup$
    – rschwieb
    Feb 27, 2018 at 12:19

1 Answer 1


Let $S$ and $T$ be minimal isomorphic ideals. As you observed, they have the same annihilator.

By Brauer's lemma, $S=eR$ for some idempotent $e$, and $T=fR$ for some idempotent $f$. (The case of $S^2=\{0\}$ is ruled out by reducedness, you see.)

Since $ann(eR)=(1-e)R$ and $ann(fR)=(1-f)R$, we see that $(1-e)R=(1-f)R$.

From $1-e=(1-f)r$, we can multiply both sides with $f$ to get $f=ef$. Doing the same thing for $1-f=(1-e)s$, we learn $e=ef$. Thus $f=e$ and $S=T$. So distinct minimal ideals cannot be isomorphic.


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